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The natural molar abundance of \(^{13} \mathrm{C}\) is roughly \(1 \%\) What is the probability of having a single \(^{13} \mathrm{C}\) isotope in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right) ?\) What is the probability that two \(^{13} \mathrm{C}\) isotopes will be adjacent to each other in benzene?

Short Answer

Expert verified
The probability of having a single \(^{13}\mathrm{C}\) isotope in benzene is approximately 5.73%, and the probability of having two \(^{13}\mathrm{C}\) isotopes adjacent to each other is approximately 0.06%.

Step by step solution

01

Calculate the probability of having a single 13C isotope in benzene

To calculate the probability of having a single 13C isotope in benzene, we need to remember that there are 6 carbon atoms in benzene and the natural molar abundance of 13C is about 1%. Therefore, the probability of having a single 13C isotope in one position in benzene is 0.01. Since there are 6 positions for carbon atoms, we can calculate the probability of having a single 13C isotope in benzene using binomial distribution: \(P(\text{single}\, ^{13}\mathrm{C}) = C_{6}^1(0.01)^1(0.99)^5 \)
02

Calculate the binomial coefficient C6^1

The binomial coefficient C6^1 can be calculated as follows: \(C_{6}^1 = \frac{6!}{1!(6-1)!} = \frac{6}{1} = 6 \)
03

Calculate the probability of having a single 13C isotope

Now we can plug in the values into the formula: \(P(\text{single}\, ^{13}\mathrm{C}) = 6(0.01)^1(0.99)^5 \approx 0.0573 \) So, the probability of having a single 13C isotope in benzene is approximately 5.73%.
04

Calculate the probability of having two 13C isotopes adjacent to each other

There are 6 possible ways for the two 13C isotopes to be adjacent in benzene (we can treat the structure as cyclical): 1. Positions 1 and 2 2. Positions 2 and 3 3. Positions 3 and 4 4. Positions 4 and 5 5. Positions 5 and 6 6. Positions 6 and 1 The probability of having two 13C isotopes adjacent to each other in any of these ways is 0.01^2, so the total probability of having two 13C isotopes adjacent to each other can be obtained by multiplying the probability by the total number of ways (6): \(P(\text{two adjacent}\, ^{13}\mathrm{C}) = 6(0.01)^2 \approx 0.0006 \) So, the probability of having two 13C isotopes adjacent to each other in benzene is approximately 0.06%. To conclude, the probability of having a single 13C isotope in benzene is approximately 5.73%, and the probability of having two 13C isotopes adjacent to each other is approximately 0.06%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Abundance
Molar abundance represents how common a particular isotope is relative to all isotopes of that element. With isotopes, each element can have different versions; for instance, carbon can be found as either \( ^{12}\mathrm{C} \) or \( ^{13}\mathrm{C} \). The natural molar abundance of \( ^{13}\mathrm{C} \) means that in a typical sample of carbon, only about 1% would be \( ^{13}\mathrm{C} \).
This small percentage plays a crucial role in calculations involving isotopes since it determines the likelihood, or probability, of their presence in compounds such as benzene \( (\mathrm{C}_6\mathrm{H}_6) \). Knowing how often an isotope naturally occurs helps in predicting its effects on the compound's properties.
  • For probabilistic calculations, consider the molar abundance as the probability of a particular isotope appearing in one atom of an element.
Isotopes
Isotopes refer to different forms of the same element that have the same number of protons but a different number of neutrons. This causes a difference in their atomic mass. For example, \( ^{12}\mathrm{C} \) and \( ^{13}\mathrm{C} \) are isotopes of carbon. They both have 6 protons, but \( ^{12}\mathrm{C} \) has 6 neutrons and \( ^{13}\mathrm{C} \) has 7 neutrons.
Even though isotopes have slightly different physical properties due to their mass difference, they typically behave very similarly in chemical reactions. This allows us to use isotopes as a scientific tool for various analyses, such as dating archaeological finds or tracing chemical pathways. In the context of the problem, the presence of \( ^{13}\mathrm{C} \) isotopes in benzene could slightly alter its mass or its reactivity under certain conditions. However, the main focus is on calculating their probability of occurrence.
  • The isotope in our calculation, \( ^{13}\mathrm{C} \), affects both how often it appears in a carbon structure and how it positions itself, as seen in the probability question of adjacency in benzene.
Probability Calculation
Probability calculations for isotopes in compounds like benzene involve understanding how likely it is for a particular arrangement of atoms to occur. This involves using concepts from binomial distributions. The probability of finding a single \( ^{13}\mathrm{C} \) in benzene uses the binomial coefficient, which counts the number of ways the isotope can be positioned.
Given that benzene has 6 carbon atoms and each has a 1% chance of being \( ^{13}\mathrm{C} \), the probability of a single \( ^{13}\mathrm{C} \) is calculated using the formula:
\[P(\text{single } ^{13}\mathrm{C}) = C_6^1 (0.01)^1 (0.99)^5\]
  • This approach considers both the chance to pick one \( ^{13}\mathrm{C} \) and how the remaining carbon atoms are likely \( ^{12}\mathrm{C} \).
  • For two adjacent \( ^{13}\mathrm{C} \) isotopes, the calculation accounts for cyclical arrangements in the benzene ring, multiplying a similar small probability by the number of possible adjacent positions.
Thus, careful consideration of these probabilities and plentiful knowledge of interesting arrangements lead to comprehensive understanding of the compound behavior.

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